OP Malhotra Class 10 Maths Solutions Chapter 5 Quadratic Equations Exercise 5 (C)

S Chand Class 10 ICSE Maths Solutions Chapter 5 Quadratic Equations Ex 5(c)

Using quadratic formula, find the roots of the following equations:

 

Question 1. \( 2x^2 + x - 3 = 0 \)
Answer: The given quadratic equation is \( 2x^2 + x - 3 = 0 \).
First, we identify the coefficients: \( a = 2 \), \( b = 1 \), and \( c = -3 \).
Next, we calculate the discriminant \( D \), which is \( b^2 - 4ac \).
\( D = (1)^2 - 4(2)(-3) \)
\( D = 1 + 24 = 25 \)
Now, we use the quadratic formula to find the roots: \( x = \frac{-b \pm \sqrt{D}}{2a} \).
\( x = \frac{-1 \pm \sqrt{25}}{2 \times 2} \)
\( x = \frac{-1 \pm 5}{4} \)
We find two roots from this:
\( x_1 = \frac{-1+5}{4} = \frac{4}{4} = 1 \)
\( x_2 = \frac{-1-5}{4} = \frac{-6}{4} = -\frac{3}{2} \)
Thus, the roots of the equation are \( x = 1 \) and \( x = -\frac{3}{2} \). The discriminant being a perfect square means the roots are rational.
In simple words: For the equation \( 2x^2 + x - 3 = 0 \), we find that \( x \) can be either 1 or negative three-halves. These are the two answers that make the equation true.

🎯 Exam Tip: Always double-check your arithmetic, especially when calculating the discriminant, as an error here will lead to incorrect roots. Remember that a positive discriminant indicates two real and distinct roots.

 

Question 2. \( 6x^2 + 7x - 20 = 0 \)
Answer: The given quadratic equation is \( 6x^2 + 7x - 20 = 0 \).
Here, the coefficients are: \( a = 6 \), \( b = 7 \), and \( c = -20 \).
Let's compute the discriminant \( D = b^2 - 4ac \).
\( D = (7)^2 - 4(6)(-20) \)
\( D = 49 + 480 = 529 \)
Now, apply the quadratic formula: \( x = \frac{-b \pm \sqrt{D}}{2a} \).
\( x = \frac{-7 \pm \sqrt{529}}{2 \times 6} \)
\( x = \frac{-7 \pm 23}{12} \)
This gives us two distinct roots:
\( x_1 = \frac{-7+23}{12} = \frac{16}{12} = \frac{4}{3} \)
\( x_2 = \frac{-7-23}{12} = \frac{-30}{12} = -\frac{5}{2} \)
So, the roots are \( x = \frac{4}{3} \) or \( x = 1\frac{1}{3} \), and \( x = -\frac{5}{2} \) or \( x = -2\frac{1}{2} \). The square root of 529 is 23, which is a prime number.
In simple words: For the equation \( 6x^2 + 7x - 20 = 0 \), the answers for \( x \) are four-thirds and negative five-halves. These are the numbers that make the equation true.

🎯 Exam Tip: When simplifying fractions in your roots, make sure to divide both the numerator and denominator by their greatest common divisor to get the simplest form.

 

Question 3. \( 9x^2 + 6x = 35 \)
Answer: The given equation is \( 9x^2 + 6x = 35 \).
First, we rewrite it in the standard quadratic form \( ax^2 + bx + c = 0 \):
\( 9x^2 + 6x - 35 = 0 \)
The coefficients are: \( a = 9 \), \( b = 6 \), and \( c = -35 \).
Calculate the discriminant \( D = b^2 - 4ac \).
\( D = (6)^2 - 4(9)(-35) \)
\( D = 36 + 1260 = 1296 \)
Now, use the quadratic formula: \( x = \frac{-b \pm \sqrt{D}}{2a} \).
\( x = \frac{-6 \pm \sqrt{1296}}{2 \times 9} \)
\( x = \frac{-6 \pm 36}{18} \)
Let's find the two roots:
\( x_1 = \frac{-6+36}{18} = \frac{30}{18} = \frac{5}{3} = 1\frac{2}{3} \)
\( x_2 = \frac{-6-36}{18} = \frac{-42}{18} = -\frac{7}{3} = -2\frac{1}{3} \)
So, the roots are \( x = 1\frac{2}{3} \) and \( x = -2\frac{1}{3} \). Finding the square root of 1296 is crucial for this problem.
In simple words: For the equation \( 9x^2 + 6x = 35 \), the possible values for \( x \) are one and two-thirds, and negative two and one-third.

🎯 Exam Tip: Always rearrange the equation into standard form \( ax^2 + bx + c = 0 \) before identifying coefficients `\(a\)`, `\(b\)`, and `\(c\)` to avoid sign errors.

 

Question 4. \( 3x^2 + 7x - 6 = 0 \)
Answer: The given quadratic equation is \( 3x^2 + 7x - 6 = 0 \).
The coefficients are: \( a = 3 \), \( b = 7 \), and \( c = -6 \).
Let's find the discriminant \( D = b^2 - 4ac \).
\( D = (7)^2 - 4(3)(-6) \)
\( D = 49 + 72 = 121 \)
Using the quadratic formula \( x = \frac{-b \pm \sqrt{D}}{2a} \):
\( x = \frac{-7 \pm \sqrt{121}}{2 \times 3} \)
\( x = \frac{-7 \pm 11}{6} \)
The two roots are:
\( x_1 = \frac{-7+11}{6} = \frac{4}{6} = \frac{2}{3} \)
\( x_2 = \frac{-7-11}{6} = \frac{-18}{6} = -3 \)
Thus, the roots of the equation are \( x = \frac{2}{3} \) and \( x = -3 \). The number 121 is a perfect square, making the roots easy to find.
In simple words: For the equation \( 3x^2 + 7x - 6 = 0 \), the answers for \( x \) are two-thirds and negative 3.

🎯 Exam Tip: Remember that `\(\sqrt{121}\)` is 11. Knowing common perfect squares can speed up calculations and reduce errors.

 

Question 5. \( x^2 – 66x + 189 = 0 \)
Answer: The given quadratic equation is \( x^2 - 66x + 189 = 0 \).
The coefficients are: \( a = 1 \), \( b = -66 \), and \( c = 189 \).
Calculate the discriminant \( D = b^2 - 4ac \).
\( D = (-66)^2 - 4(1)(189) \)
\( D = 4356 - 756 = 3600 \)
Now, use the quadratic formula: \( x = \frac{-b \pm \sqrt{D}}{2a} \).
\( x = \frac{-(-66) \pm \sqrt{3600}}{2 \times 1} \)
\( x = \frac{66 \pm 60}{2} \)
Let's find the two roots:
\( x_1 = \frac{66+60}{2} = \frac{126}{2} = 63 \)
\( x_2 = \frac{66-60}{2} = \frac{6}{2} = 3 \)
So, the roots of the equation are \( x = 63 \) and \( x = 3 \). When the discriminant is a perfect square, the roots are always rational numbers.
In simple words: For the equation \( x^2 - 66x + 189 = 0 \), the values that \( x \) can be are 63 and 3.

🎯 Exam Tip: Recognizing that a large discriminant like 3600 is a perfect square (e.g., \(60^2\)) can simplify your calculations significantly.

 

Question 6. \( \sqrt{3}x^2 + 11x + 6\sqrt{3} = 0 \)
Answer: The given quadratic equation is \( \sqrt{3}x^2 + 11x + 6\sqrt{3} = 0 \).
The coefficients are: \( a = \sqrt{3} \), \( b = 11 \), and \( c = 6\sqrt{3} \).
Calculate the discriminant \( D = b^2 - 4ac \).
\( D = (11)^2 - 4(\sqrt{3})(6\sqrt{3}) \)
\( D = 121 - 4(6)(3) \)
\( D = 121 - 72 = 49 \)
Now, apply the quadratic formula: \( x = \frac{-b \pm \sqrt{D}}{2a} \).
\( x = \frac{-11 \pm \sqrt{49}}{2 \times \sqrt{3}} \)
\( x = \frac{-11 \pm 7}{2\sqrt{3}} \)
We find the two roots:
\( x_1 = \frac{-11+7}{2\sqrt{3}} = \frac{-4}{2\sqrt{3}} = \frac{-2}{\sqrt{3}} \)
To rationalize the denominator:
\( x_1 = \frac{-2\sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{-2\sqrt{3}}{3} \)
For the second root:
\( x_2 = \frac{-11-7}{2\sqrt{3}} = \frac{-18}{2\sqrt{3}} = \frac{-9}{\sqrt{3}} \)
To rationalize the denominator:
\( x_2 = \frac{-9\sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{-9\sqrt{3}}{3} = -3\sqrt{3} \)
So, the roots are \( x = \frac{-2\sqrt{3}}{3} \) and \( x = -3\sqrt{3} \). Rationalizing the denominator is a common step when dealing with square roots in the solution.
In simple words: For the equation \( \sqrt{3}x^2 + 11x + 6\sqrt{3} = 0 \), the two answers for \( x \) are negative two-thirds of root 3, and negative 3 times root 3.

🎯 Exam Tip: When working with square roots in the denominator, always remember to rationalize the denominator by multiplying both the numerator and denominator by the radical term.

 

Question 7. \( 36x^2 + 23 = 60x \)
Answer: The given equation is \( 36x^2 + 23 = 60x \).
First, rearrange it into the standard quadratic form \( ax^2 + bx + c = 0 \):
\( 36x^2 - 60x + 23 = 0 \)
The coefficients are: \( a = 36 \), \( b = -60 \), and \( c = 23 \).
Calculate the discriminant \( D = b^2 - 4ac \).
\( D = (-60)^2 - 4(36)(23) \)
\( D = 3600 - 3312 = 288 \)
Now, apply the quadratic formula: \( x = \frac{-b \pm \sqrt{D}}{2a} \).
\( x = \frac{-(-60) \pm \sqrt{288}}{2 \times 36} \)
\( x = \frac{60 \pm \sqrt{144 \times 2}}{72} \)
\( x = \frac{60 \pm 12\sqrt{2}}{72} \)
To simplify, divide the numerator and denominator by their common factor, 12:
\( x = \frac{5 \pm \sqrt{2}}{6} \)
So, the roots are \( x = \frac{5+\sqrt{2}}{6} \) and \( x = \frac{5-\sqrt{2}}{6} \). Simplifying the square root of 288 to \(12\sqrt{2}\) helps in further simplification of the fraction.
In simple words: For the equation \( 36x^2 + 23 = 60x \), the answers for \( x \) are five plus root 2, all divided by 6, and five minus root 2, all divided by 6.

🎯 Exam Tip: Always simplify the radical in the discriminant to its simplest mixed radical form (e.g., `\(\sqrt{288} = 12\sqrt{2}\)`) before attempting to simplify the entire fraction.

 

Question 8. \( x^2 - 2x + 5 = 0 \)
Answer: The given quadratic equation is \( x^2 - 2x + 5 = 0 \).
The coefficients are: \( a = 1 \), \( b = -2 \), and \( c = 5 \).
Calculate the discriminant \( D = b^2 - 4ac \).
\( D = (-2)^2 - 4(1)(5) \)
\( D = 4 - 20 = -16 \)
Now, use the quadratic formula: \( x = \frac{-b \pm \sqrt{D}}{2a} \).
\( x = \frac{-(-2) \pm \sqrt{-16}}{2 \times 1} \)
\( x = \frac{2 \pm \sqrt{16 \times (-1)}}{2} \)
\( x = \frac{2 \pm 4\sqrt{-1}}{2} \)
Since `\(\sqrt{-1}\)` is represented by `\(i\)` (the imaginary unit), we have:
\( x = \frac{2 \pm 4i}{2} \)
\( x = 1 \pm 2i \)
So, the roots are \( x = 1 + 2i \) and \( x = 1 - 2i \). Since the discriminant is negative, these are not real roots; they are complex numbers. When the discriminant is negative, the quadratic equation has complex roots, which involve the imaginary unit `\(i\)`.
In simple words: For the equation \( x^2 - 2x + 5 = 0 \), the answers for \( x \) are 1 plus 2i, and 1 minus 2i. These are not ordinary numbers but 'complex' numbers because they involve `\(i\)` (imaginary number).

🎯 Exam Tip: If the discriminant `\(D\)` is negative, the equation has no real roots. The roots are complex conjugates, involving the imaginary unit `\(i = \sqrt{-1}\)`. Ensure you write them in the form `\(a \pm bi\)`.

 

Question 9. \( 3x^2 - 17x + 25 = 0 \)
Answer: The given quadratic equation is \( 3x^2 - 17x + 25 = 0 \).
The coefficients are: \( a = 3 \), \( b = -17 \), and \( c = 25 \).
Calculate the discriminant \( D = b^2 - 4ac \).
\( D = (-17)^2 - 4(3)(25) \)
\( D = 289 - 300 = -11 \)
Now, use the quadratic formula: \( x = \frac{-b \pm \sqrt{D}}{2a} \).
\( x = \frac{-(-17) \pm \sqrt{-11}}{2 \times 3} \)
\( x = \frac{17 \pm \sqrt{-11}}{6} \)
Since `\(\sqrt{-1}\)` is `\(i\)`, we can write this as:
\( x = \frac{17 \pm i\sqrt{11}}{6} \)
So, the roots are \( x = \frac{17 + i\sqrt{11}}{6} \) and \( x = \frac{17 - i\sqrt{11}}{6} \). Again, these are not real roots, but complex roots. For quadratic equations with a negative discriminant, the solutions always involve imaginary numbers.
In simple words: For the equation \( 3x^2 - 17x + 25 = 0 \), the answers for \( x \) are complex. They are 17 plus imaginary root 11, all divided by 6, and 17 minus imaginary root 11, all divided by 6.

🎯 Exam Tip: When presenting complex roots, remember that `\(\sqrt{-k} = i\sqrt{k}\)` for a positive number `\(k\)`. This notation clearly separates the real and imaginary parts.

 

Question 10. \( 15x^2 - 28 = x \)
Answer: The given equation is \( 15x^2 - 28 = x \).
First, rewrite it in the standard quadratic form \( ax^2 + bx + c = 0 \):
\( 15x^2 - x - 28 = 0 \)
The coefficients are: \( a = 15 \), \( b = -1 \), and \( c = -28 \).
Calculate the discriminant \( D = b^2 - 4ac \).
\( D = (-1)^2 - 4(15)(-28) \)
\( D = 1 + 1680 = 1681 \)
Now, apply the quadratic formula: \( x = \frac{-b \pm \sqrt{D}}{2a} \).
\( x = \frac{-(-1) \pm \sqrt{1681}}{2 \times 15} \)
\( x = \frac{1 \pm 41}{30} \)
Let's find the two roots:
\( x_1 = \frac{1+41}{30} = \frac{42}{30} = \frac{7}{5} = 1\frac{2}{5} \)
\( x_2 = \frac{1-41}{30} = \frac{-40}{30} = -\frac{4}{3} = -1\frac{1}{3} \)
So, the roots are \( x = 1\frac{2}{5} \) and \( x = -1\frac{1}{3} \). The square root of 1681 is 41, which helps in getting integer values for the numerator before simplification.
In simple words: For the equation \( 15x^2 - 28 = x \), the answers for \( x \) are one and two-fifths, and negative one and one-third.

🎯 Exam Tip: It is helpful to memorize squares of numbers up to at least 50 (like \(41^2 = 1681\)) to quickly identify perfect squares for the discriminant.

 

Question 11. x² + 3x − 3 = 0, giving your answer to two decimal places.
Answer: The given quadratic equation is \( x^2 + 3x - 3 = 0 \).
The coefficients are: \( a = 1 \), \( b = 3 \), and \( c = -3 \).
Calculate the discriminant \( D = b^2 - 4ac \).
\( D = (3)^2 - 4(1)(-3) \)
\( D = 9 + 12 = 21 \)
Now, use the quadratic formula: \( x = \frac{-b \pm \sqrt{D}}{2a} \).
\( x = \frac{-3 \pm \sqrt{21}}{2 \times 1} \)
To get the answer to two decimal places, we approximate `\(\sqrt{21}\)`: `\(\sqrt{21} \approx 4.5826 \approx 4.58\)` (rounded to two decimal places).
\( x = \frac{-3 \pm 4.58}{2} \)
Let's find the two roots:
\( x_1 = \frac{-3+4.58}{2} = \frac{1.58}{2} = 0.79 \)
\( x_2 = \frac{-3-4.58}{2} = \frac{-7.58}{2} = -3.79 \)
So, the roots of the equation, rounded to two decimal places, are \( x = 0.79 \) and \( x = -3.79 \). When asked for answers to a specific decimal place, make sure to round your final answers accurately.
In simple words: For the equation \( x^2 + 3x - 3 = 0 \), the answers for \( x \), rounded to two decimal places, are 0.79 and negative 3.79.

🎯 Exam Tip: When rounding, always carry one extra decimal place in your intermediate calculations to ensure accuracy in the final rounded answer.

 

Question 12. \( \frac { 2 }{ 3 }x = \frac { -1 }{ 6 }x^2 – \frac { 1 }{ 3 } \), giving your answer to 2 d.p.
Answer: The given equation is \( \frac{2}{3}x = -\frac{1}{6}x^2 - \frac{1}{3} \).
First, rearrange it into the standard quadratic form \( ax^2 + bx + c = 0 \):
\( \frac{1}{6}x^2 + \frac{2}{3}x + \frac{1}{3} = 0 \)
To eliminate the fractions, multiply the entire equation by the Least Common Multiple (LCM) of the denominators (6, 3, 3), which is 6:
\( 6 \times (\frac{1}{6}x^2 + \frac{2}{3}x + \frac{1}{3}) = 6 \times 0 \)
\( x^2 + 4x + 2 = 0 \)
The coefficients are: \( a = 1 \), \( b = 4 \), and \( c = 2 \).
Calculate the discriminant \( D = b^2 - 4ac \).
\( D = (4)^2 - 4(1)(2) \)
\( D = 16 - 8 = 8 \)
Now, use the quadratic formula: \( x = \frac{-b \pm \sqrt{D}}{2a} \).
\( x = \frac{-4 \pm \sqrt{8}}{2 \times 1} \)
\( x = \frac{-4 \pm 2\sqrt{2}}{2} \)
Divide by 2:
\( x = -2 \pm \sqrt{2} \)
To get the answer to two decimal places, we approximate `\(\sqrt{2}\)`: `\(\sqrt{2} \approx 1.414 \approx 1.41\)` (rounded to two decimal places).
\( x = -2 \pm 1.41 \)
Let's find the two roots:
\( x_1 = -2 + 1.41 = -0.59 \)
\( x_2 = -2 - 1.41 = -3.41 \)
So, the roots of the equation, rounded to two decimal places, are \( x = -0.59 \) and \( x = -3.41 \). Multiplying by the Least Common Multiple (LCM) of the denominators helps convert the fractional equation into a standard quadratic form.
In simple words: For the equation \( \frac{2}{3}x = -\frac{1}{6}x^2 - \frac{1}{3} \), the answers for \( x \), rounded to two decimal places, are negative 0.59 and negative 3.41.

🎯 Exam Tip: Always clear fractions from an equation by multiplying by the LCM of denominators. This simplifies calculations and reduces the chance of errors.

 

Question 13. x² + 6x - 10 = 0
Answer: The given quadratic equation is \( x^2 + 6x - 10 = 0 \).
The coefficients are: \( a = 1 \), \( b = 6 \), and \( c = -10 \).
Calculate the discriminant \( D = b^2 - 4ac \).
\( D = (6)^2 - 4(1)(-10) \)
\( D = 36 + 40 = 76 \)
Now, use the quadratic formula: \( x = \frac{-b \pm \sqrt{D}}{2a} \).
\( x = \frac{-6 \pm \sqrt{76}}{2 \times 1} \)
To simplify `\(\sqrt{76}\)`, we find the largest perfect square factor:
\( \sqrt{76} = \sqrt{4 \times 19} = 2\sqrt{19} \)
So, the roots are:
\( x = \frac{-6 \pm 2\sqrt{19}}{2} \)
Divide the numerator and denominator by 2:
\( x = -3 \pm \sqrt{19} \)
Thus, the roots are \( x = -3 + \sqrt{19} \) and \( x = -3 - \sqrt{19} \). When the discriminant is not a perfect square, the roots are irrational and are often left in simplest radical form.
In simple words: For the equation \( x^2 + 6x - 10 = 0 \), the answers for \( x \) are negative 3 plus root 19, and negative 3 minus root 19.

🎯 Exam Tip: Always simplify square roots in your final answer. Look for perfect square factors within the radical to extract them (e.g., `\(\sqrt{76} = \sqrt{4 \times 19} = 2\sqrt{19}\)`).

 

Question 14. \( \frac{x^2+8}{11} = 5x – x^2 – 5 \)
Answer: The given equation is \( \frac{x^2+8}{11} = 5x - x^2 - 5 \).
First, multiply both sides by 11 to clear the denominator:
\( 11 \times \frac{x^2+8}{11} = 11 \times (5x - x^2 - 5) \)
\( x^2 + 8 = 55x - 11x^2 - 55 \)
Next, rearrange the equation into the standard quadratic form \( ax^2 + bx + c = 0 \):
\( x^2 + 11x^2 - 55x + 8 + 55 = 0 \)
\( 12x^2 - 55x + 63 = 0 \)
The coefficients are: \( a = 12 \), \( b = -55 \), and \( c = 63 \).
Calculate the discriminant \( D = b^2 - 4ac \).
\( D = (-55)^2 - 4(12)(63) \)
\( D = 3025 - 3024 = 1 \)
Now, apply the quadratic formula: \( x = \frac{-b \pm \sqrt{D}}{2a} \).
\( x = \frac{-(-55) \pm \sqrt{1}}{2 \times 12} \)
\( x = \frac{55 \pm 1}{24} \)
Let's find the two roots:
\( x_1 = \frac{55+1}{24} = \frac{56}{24} = \frac{7}{3} = 2\frac{1}{3} \)
\( x_2 = \frac{55-1}{24} = \frac{54}{24} = \frac{9}{4} = 2\frac{1}{4} \)
So, the roots are \( x = 2\frac{1}{3} \) and \( x = 2\frac{1}{4} \). The value of the discriminant being 1 (a perfect square) means the roots are rational and distinct.
In simple words: For the equation \( \frac{x^2+8}{11} = 5x - x^2 - 5 \), the answers for \( x \) are two and one-third, and two and one-fourth.

🎯 Exam Tip: When dealing with fractional equations, multiply by the denominator to clear fractions and simplify the equation into standard quadratic form before applying the formula.

 

Question 15. \( y - \frac { 3 }{ y } = \frac { 1 }{ 2 } \)
Answer: The given equation is \( y - \frac{3}{y} = \frac{1}{2} \).
To eliminate the denominators, multiply the entire equation by the LCM of the denominators (`\(y\)` and `\(2\)`), which is `\(2y\)`:
\( 2y \times (y - \frac{3}{y}) = 2y \times (\frac{1}{2}) \)
\( 2y^2 - 6 = y \)
Rearrange the equation into the standard quadratic form \( ay^2 + by + c = 0 \):
\( 2y^2 - y - 6 = 0 \)
The coefficients are: \( a = 2 \), \( b = -1 \), and \( c = -6 \).
Calculate the discriminant \( D = b^2 - 4ac \).
\( D = (-1)^2 - 4(2)(-6) \)
\( D = 1 + 48 = 49 \)
Now, apply the quadratic formula: \( y = \frac{-b \pm \sqrt{D}}{2a} \).
\( y = \frac{-(-1) \pm \sqrt{49}}{2 \times 2} \)
\( y = \frac{1 \pm 7}{4} \)
Let's find the two roots:
\( y_1 = \frac{1+7}{4} = \frac{8}{4} = 2 \)
\( y_2 = \frac{1-7}{4} = \frac{-6}{4} = -\frac{3}{2} = -1\frac{1}{2} \)
So, the roots are \( y = 2 \) and \( y = -1\frac{1}{2} \). Always check for extraneous roots when solving rational equations by substituting the solutions back into the original equation.
In simple words: For the equation \( y - \frac{3}{y} = \frac{1}{2} \), the answers for \( y \) are 2 and negative one and a half.

🎯 Exam Tip: When the variable is in the denominator, remember that the variable cannot be zero. Any solution that results in a zero denominator for the original equation must be rejected.

 

Question 16. \( 2x + \frac { 4 }{ x } = 9 \)
Answer: The given equation is \( 2x + \frac{4}{x} = 9 \).
To eliminate the denominator, multiply the entire equation by `\(x\)` (assuming \( x \ne 0 \)):
\( x \times (2x + \frac{4}{x}) = x \times 9 \)
\( 2x^2 + 4 = 9x \)
Rearrange the equation into the standard quadratic form \( ax^2 + bx + c = 0 \):
\( 2x^2 - 9x + 4 = 0 \)
The coefficients are: \( a = 2 \), \( b = -9 \), and \( c = 4 \).
Calculate the discriminant \( D = b^2 - 4ac \).
\( D = (-9)^2 - 4(2)(4) \)
\( D = 81 - 32 = 49 \)
Now, apply the quadratic formula: \( x = \frac{-b \pm \sqrt{D}}{2a} \).
\( x = \frac{-(-9) \pm \sqrt{49}}{2 \times 2} \)
\( x = \frac{9 \pm 7}{4} \)
Let's find the two roots:
\( x_1 = \frac{9+7}{4} = \frac{16}{4} = 4 \)
\( x_2 = \frac{9-7}{4} = \frac{2}{4} = \frac{1}{2} \)
So, the roots are \( x = 4 \) and \( x = \frac{1}{2} \). This type of equation is transformed into a quadratic by multiplying all terms by `\(x\)`, ensuring `\(x \ne 0\)`.
In simple words: For the equation \( 2x + \frac{4}{x} = 9 \), the answers for \( x \) are 4 and one-half.

🎯 Exam Tip: For equations with variables in the denominator, always state restrictions on the variable (e.g., `\(x \ne 0\)` in this case) to ensure that your solutions are valid within the domain of the original equation.

 

Question 17. \( \frac{x}{x+1}+\frac{x+1}{x}=\frac{34}{15} \), where \( x \ne 0 \), \( x \ne -1 \)
Answer: The given equation is \( \frac{x}{x+1}+\frac{x+1}{x}=\frac{34}{15} \).
First, combine the fractions on the left-hand side:
\( \frac{x(x) + (x+1)(x+1)}{x(x+1)} = \frac{34}{15} \)
\( \frac{x^2 + (x^2+2x+1)}{x^2+x} = \frac{34}{15} \)
\( \frac{2x^2+2x+1}{x^2+x} = \frac{34}{15} \)
Now, cross-multiply:
\( 15(2x^2+2x+1) = 34(x^2+x) \)
\( 30x^2+30x+15 = 34x^2+34x \)
Rearrange the terms to get the standard quadratic form \( ax^2 + bx + c = 0 \):
\( 0 = 34x^2 - 30x^2 + 34x - 30x - 15 \)
\( 4x^2 + 4x - 15 = 0 \)
The coefficients are: \( a = 4 \), \( b = 4 \), and \( c = -15 \).
Calculate the discriminant \( D = b^2 - 4ac \).
\( D = (4)^2 - 4(4)(-15) \)
\( D = 16 + 240 = 256 \)
Now, apply the quadratic formula: \( x = \frac{-b \pm \sqrt{D}}{2a} \).
\( x = \frac{-4 \pm \sqrt{256}}{2 \times 4} \)
\( x = \frac{-4 \pm 16}{8} \)
Let's find the two roots:
\( x_1 = \frac{-4+16}{8} = \frac{12}{8} = \frac{3}{2} = 1\frac{1}{2} \)
\( x_2 = \frac{-4-16}{8} = \frac{-20}{8} = -\frac{5}{2} = -2\frac{1}{2} \)
So, the roots are \( x = 1\frac{1}{2} \) and \( x = -2\frac{1}{2} \). Remember to identify the domain of the variable (values \(x\) cannot be) before solving, to ensure the solutions are valid.
In simple words: For the equation \( \frac{x}{x+1}+\frac{x+1}{x}=\frac{34}{15} \), the answers for \( x \) are one and a half, and negative two and a half.

🎯 Exam Tip: Always note the values of `\(x\)` that would make any denominator zero in the original equation (e.g., \(x \ne 0\), \(x \ne -1\)). These values are excluded from the solution set.

 

Question 18. \( \frac{2x}{x-4}+\frac{2x-5}{x-3}=8\frac{1}{3} \)
Answer: The given equation is \( \frac{2x}{x-4}+\frac{2x-5}{x-3}=8\frac{1}{3} \).
First, convert the mixed fraction to an improper fraction: \( 8\frac{1}{3} = \frac{25}{3} \).
Now, combine the fractions on the left-hand side:
\( \frac{2x(x-3) + (2x-5)(x-4)}{(x-4)(x-3)} = \frac{25}{3} \)
Expand the numerator and denominator:
\( \frac{2x^2-6x + (2x^2-8x-5x+20)}{x^2-3x-4x+12} = \frac{25}{3} \)
\( \frac{2x^2-6x + 2x^2-13x+20}{x^2-7x+12} = \frac{25}{3} \)
\( \frac{4x^2-19x+20}{x^2-7x+12} = \frac{25}{3} \)
Next, cross-multiply:
\( 3(4x^2-19x+20) = 25(x^2-7x+12) \)
\( 12x^2-57x+60 = 25x^2-175x+300 \)
Rearrange the terms to get the standard quadratic form \( ax^2 + bx + c = 0 \):
\( 0 = 25x^2 - 12x^2 - 175x + 57x + 300 - 60 \)
\( 13x^2 - 118x + 240 = 0 \)
The coefficients are: \( a = 13 \), \( b = -118 \), and \( c = 240 \).
Calculate the discriminant \( D = b^2 - 4ac \).
\( D = (-118)^2 - 4(13)(240) \)
\( D = 13924 - 12480 = 1444 \)
Now, apply the quadratic formula: \( x = \frac{-b \pm \sqrt{D}}{2a} \).
\( x = \frac{-(-118) \pm \sqrt{1444}}{2 \times 13} \)
\( x = \frac{118 \pm 38}{26} \)
Let's find the two roots:
\( x_1 = \frac{118+38}{26} = \frac{156}{26} = 6 \)
\( x_2 = \frac{118-38}{26} = \frac{80}{26} = \frac{40}{13} = 3\frac{1}{13} \)
So, the roots are \( x = 6 \) and \( x = 3\frac{1}{13} \). Remember to specify any values of \(x\) that would make the denominators zero (in this case \(x \ne 4\) and \(x \ne 3\)), ensuring your solutions are valid.
In simple words: For the equation \( \frac{2x}{x-4}+\frac{2x-5}{x-3}=8\frac{1}{3} \), the answers for \( x \) are 6 and three and one-thirteenth.

🎯 Exam Tip: Always make sure to check the original equation for values of `\(x\)` that would make any denominator zero; these values are called extraneous solutions and must be excluded.

 

Question 19. \( \frac{x+6}{x+7}-\frac{x+1}{x+2}=\frac{1}{3 x+1} \)
Answer: The given equation is \( \frac{x+6}{x+7}-\frac{x+1}{x+2}=\frac{1}{3 x+1} \).
First, combine the fractions on the left-hand side:
\( \frac{(x+6)(x+2) - (x+1)(x+7)}{(x+7)(x+2)} = \frac{1}{3x+1} \)
Expand the products in the numerator and denominator:
\( \frac{(x^2+2x+6x+12) - (x^2+7x+x+7)}{x^2+2x+7x+14} = \frac{1}{3x+1} \)
\( \frac{(x^2+8x+12) - (x^2+8x+7)}{x^2+9x+14} = \frac{1}{3x+1} \)
Simplify the numerator:
\( \frac{x^2+8x+12-x^2-8x-7}{x^2+9x+14} = \frac{1}{3x+1} \)
\( \frac{5}{x^2+9x+14} = \frac{1}{3x+1} \)
Now, cross-multiply:
\( 5(3x+1) = 1(x^2+9x+14) \)
\( 15x+5 = x^2+9x+14 \)
Rearrange the terms to get the standard quadratic form \( ax^2 + bx + c = 0 \):
\( 0 = x^2+9x-15x+14-5 \)
\( x^2 - 6x + 9 = 0 \)
The coefficients are: \( a = 1 \), \( b = -6 \), and \( c = 9 \).
Calculate the discriminant \( D = b^2 - 4ac \).
\( D = (-6)^2 - 4(1)(9) \)
\( D = 36 - 36 = 0 \)
Now, apply the quadratic formula: \( x = \frac{-b \pm \sqrt{D}}{2a} \).
\( x = \frac{-(-6) \pm \sqrt{0}}{2 \times 1} \)
\( x = \frac{6 \pm 0}{2} \)
\( x = \frac{6}{2} = 3 \)
So, the only root is \( x = 3 \). A discriminant of zero means the quadratic equation has exactly one real root (a repeated root).
In simple words: For the equation \( \frac{x+6}{x+7}-\frac{x+1}{x+2}=\frac{1}{3 x+1} \), the only answer for \( x \) that makes the equation true is 3.

🎯 Exam Tip: When the discriminant is zero, the quadratic equation has one repeated real root. This means the quadratic is a perfect square trinomial.

 

Question 20. \( \frac{x+1}{2 x+5}=\frac{x+3}{3 x+4} \)
Answer: The given equation is \( \frac{x+1}{2x+5} = \frac{x+3}{3x+4} \).
First, cross-multiply the terms:
\( (x+1)(3x+4) = (x+3)(2x+5) \)
Expand both sides of the equation:
\( 3x^2+4x+3x+4 = 2x^2+5x+6x+15 \)
Combine like terms on each side:
\( 3x^2+7x+4 = 2x^2+11x+15 \)
Rearrange the equation into the standard quadratic form \( ax^2 + bx + c = 0 \):
\( 3x^2 - 2x^2 + 7x - 11x + 4 - 15 = 0 \)
\( x^2 - 4x - 11 = 0 \)
The coefficients are: \( a = 1 \), \( b = -4 \), and \( c = -11 \).
Calculate the discriminant \( D = b^2 - 4ac \).
\( D = (-4)^2 - 4(1)(-11) \)
\( D = 16 + 44 = 60 \)
Now, apply the quadratic formula: \( x = \frac{-b \pm \sqrt{D}}{2a} \).
\( x = \frac{-(-4) \pm \sqrt{60}}{2 \times 1} \)
\( x = \frac{4 \pm \sqrt{4 \times 15}}{2} \)
\( x = \frac{4 \pm 2\sqrt{15}}{2} \)
Divide the numerator by 2:
\( x = 2 \pm \sqrt{15} \)
So, the roots are \( x = 2 + \sqrt{15} \) and \( x = 2 - \sqrt{15} \). Always simplify radical terms in the solution, like `\(\sqrt{60}\)` to `\(2\sqrt{15}\)`.
In simple words: For the equation \( \frac{x+1}{2 x+5}=\frac{x+3}{3 x+4} \), the answers for \( x \) are 2 plus root 15, and 2 minus root 15.

🎯 Exam Tip: Pay close attention to expanding binomials correctly on both sides of the equation to avoid algebraic errors that can lead to incorrect coefficients.

 

Question 21. Solve using the quadratic formula:
(i) a²x² - 3abx + 2b² = 0
(ii) x² - x - a(a + 1) = 0
(iii) 10x² + 3bx + a² – 7ax – b² = 0
Answer:
(i) Given quadratic equation: \( a^2x^2 - 3abx + 2b^2 = 0 \).
Here, the coefficients are: \( A = a^2 \), \( B = -3ab \), and \( C = 2b^2 \).
First, calculate the discriminant \( D = B^2 - 4AC \).
\( D = (-3ab)^2 - 4(a^2)(2b^2) \)
\( D = 9a^2b^2 - 8a^2b^2 \)
\( D = a^2b^2 \)
Now, use the quadratic formula: \( x = \frac{-B \pm \sqrt{D}}{2A} \).
\( x = \frac{-(-3ab) \pm \sqrt{a^2b^2}}{2a^2} \)
\( x = \frac{3ab \pm ab}{2a^2} \)
We find two roots from this:
\( x_1 = \frac{3ab+ab}{2a^2} = \frac{4ab}{2a^2} = \frac{2b}{a} \)
\( x_2 = \frac{3ab-ab}{2a^2} = \frac{2ab}{2a^2} = \frac{b}{a} \)
So, the roots are \( x = \frac{2b}{a} \) and \( x = \frac{b}{a} \). This problem shows how to apply the quadratic formula even when the coefficients are algebraic expressions.

(ii) Given quadratic equation: \( x^2 - x - a(a+1) = 0 \).
First, expand the constant term: \( x^2 - x - (a^2+a) = 0 \).
Here, the coefficients are: \( A = 1 \), \( B = -1 \), and \( C = -(a^2+a) \).
Calculate the discriminant \( D = B^2 - 4AC \).
\( D = (-1)^2 - 4(1)(-(a^2+a)) \)
\( D = 1 + 4a^2 + 4a \)
\( D = (2a+1)^2 \)
Now, use the quadratic formula: \( x = \frac{-B \pm \sqrt{D}}{2A} \).
\( x = \frac{-(-1) \pm \sqrt{(2a+1)^2}}{2 \times 1} \)
\( x = \frac{1 \pm (2a+1)}{2} \)
We find two roots from this:
\( x_1 = \frac{1+(2a+1)}{2} = \frac{2a+2}{2} = a+1 \)
\( x_2 = \frac{1-(2a+1)}{2} = \frac{1-2a-1}{2} = \frac{-2a}{2} = -a \)
So, the roots are \( x = a+1 \) and \( x = -a \). Notice how \( (2a+1)^2 \) is a perfect square, making the square root straightforward and the roots rational algebraic expressions.

(iii) Given quadratic equation: \( 10x^2 + 3bx + a^2 - 7ax - b^2 = 0 \).
First, rewrite it in the standard quadratic form \( Ax^2 + Bx + C = 0 \):
\( 10x^2 + (3b-7a)x + (a^2-b^2) = 0 \)
Here, the coefficients are: \( A = 10 \), \( B = (3b-7a) \), and \( C = (a^2-b^2) \).
Calculate the discriminant \( D = B^2 - 4AC \).
\( D = (3b-7a)^2 - 4(10)(a^2-b^2) \)
\( D = (9b^2 - 42ab + 49a^2) - 40a^2 + 40b^2 \)
\( D = 9b^2 + 49a^2 - 42ab - 40a^2 + 40b^2 \)
\( D = 9a^2 - 42ab + 49b^2 \)
This can be recognized as a perfect square trinomial:
\( D = (3a-7b)^2 \)
Now, use the quadratic formula: \( x = \frac{-B \pm \sqrt{D}}{2A} \).
\( x = \frac{-(3b-7a) \pm \sqrt{(3a-7b)^2}}{2 \times 10} \)
\( x = \frac{-3b+7a \pm (3a-7b)}{20} \)
We find two roots from this:
\( x_1 = \frac{-3b+7a + (3a-7b)}{20} = \frac{10a-10b}{20} = \frac{10(a-b)}{20} = \frac{a-b}{2} \)
\( x_2 = \frac{-3b+7a - (3a-7b)}{20} = \frac{-3b+7a-3a+7b}{20} = \frac{4a+4b}{20} = \frac{4(a+b)}{20} = \frac{a+b}{5} \)
So, the roots are \( x = \frac{a-b}{2} \) and \( x = \frac{a+b}{5} \). Recognizing the discriminant as a perfect square trinomial (like \( (3a-7b)^2 \)) is key to simplifying the roots quickly.
In simple words: For (i), the answers for \( x \) are `\(2b/a\)` and `\(b/a\)`. For (ii), they are `\(a+1\)` and `\(-a\)`. For (iii), the answers are `\((a-b)/2\)` and `\((a+b)/5\)`. These problems show how to use the quadratic formula even when the numbers are replaced by letters.

🎯 Exam Tip: When dealing with algebraic coefficients, carefully group terms and factorize to simplify the discriminant, especially when it turns out to be a perfect square, making the square root step straightforward.

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